Question

find the sum of first 51 terms of an AP
whose 2nd and third terms are 14 and 18.​

Answer 1

Step-by-step explanation:

A(2) = 14 & A(3) = 18

means , Common difference (d) = 4

a + 4 = 14

a = 10

Sum of first 51 terms

S(51) = 51/2 { 2(10) + (51-1)4 }

= 51/2 { 20 + 200 }

= 51/2 { 220 }

= 51 × 110 = 51(100+10)

= 5100 + 510 = 5610

Answer 2

Step-by-step explanation:

second term:a+d=14------------(1)

third term:a+2d=18----------(2)

(2)-(1)=d=4

substitute d=4 in (1)

a+4=14

a=14-4=10

sum of first 51 terms=n/2(2a+(n-1)d)

s51=51/2(2*10+(51-1)4)

s51=51/2(20+50(4))

s51=51/2(20+200)

s51=51/2(220)

s51=51(110)

s51=5610

therefore sum of first 51 terms=5610